By exchanging the roles of $i$ and $j$, we see that $(i-j) \cdot m_k + (j-k) \cdot m_i + (k-i) \cdot m_j = C = (j-i) \cdot m_k + (i-k) \cdot m_j + (k-j) \cdot m_i = -C$ must hold, which yields $C = 0$. For $(i, j, k) = (1, 2, 3)$, we obtain
$$(1-2) \cdot \frac{a_1+a_2+a_3}{3} + (2-3) \cdot a_1 + (3-1) \cdot \frac{a_1+a_2}{2} = 0,$$
which is equivalent to
$$\frac{a_1+a_2+a_3}{3} - a_1 + a_1 + a_2 = 0 \iff a_1 + a_3 = a_2.$$
The first three elements of the sequence therefore are indeed elements of an arithmetic sequence. We can now use induction to show that the entire sequence is arithmetic, i.e. that $a_n = a_1 = (n-1)(a_2 - a_1) = (n-1)a_2 + (n-2)a_1$ holds. In order to do this, we assume that $a_k = a_1 + (k-1)(a_2 - a_1)$ holds for $1 \le k \le n-1$, and consider the triple $(i, j, k) = (1, 2, n)$. We then have
$$\begin{align*} & (1-2) \cdot \frac{\frac{(n-1)(2a_1)+(n-2)(a_2-a_1))}{2} + a_n}{n} + (2-n) \cdot a_1 + (n-1) \cdot \frac{a_1+a_2}{2} = 0 \\ \iff & \frac{a_1(3-n)+a_2(n-1)}{2} - \frac{a_1(n-1)(4-n)+a_2(n-1)(n-2)+2a_n}{2n} = 0 \\ \iff & a_1 \cdot (3n-n^2+n^2-5n+4) + a_2 \cdot (n-1)(n-n+2) = 2a_n \\ \iff & a_2 \cdot (n-1) - a_1 \cdot (n-2) = a_n, \end{align*}$$
which completes the induction. We see that the sequence is indeed arithmetic, as claimed. $\square$